package com.yubest;

/**
 * 将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。 
 *
 *  
 *
 * 示例 1：[图片] img/0021.jpg
 *
 *
 * 输入：l1 = [1,2,4], l2 = [1,3,4]
 * 输出：[1,1,2,3,4,4]
 * 示例 2：
 *
 * 输入：l1 = [], l2 = []
 * 输出：[]
 * 示例 3：
 *
 * 输入：l1 = [], l2 = [0]
 * 输出：[0]
 *  
 *
 * 提示：
 *
 * 两个链表的节点数目范围是 [0, 50]
 * -100 <= Node.val <= 100
 * l1 和 l2 均按 非递减顺序 排列
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/merge-two-sorted-lists
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 *
 * @Author hweiyu
 * @Description
 * @Date 2021/11/1 14:09
 */
public class P0021 {

    public static void main(String[] args) {
        System.out.println(new Solution21().mergeTwoLists(new ListNode(1, new ListNode(2, new ListNode(4))), new ListNode(1, new ListNode(3, new ListNode(4)))));
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution21 {

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode guard = new ListNode(-101);
        ListNode cur = guard;
        while (l1 != null || l2 != null) {
            if (null != l1 && null != l2) {
                if (l1.val > l2.val) {
                    cur.next = new ListNode(l2.val);
                    l2 = l2.next;
                } else {
                    cur.next = new ListNode(l1.val);
                    l1 = l1.next;
                }
            } else if (null != l1) {
                cur.next = new ListNode(l1.val);
                l1 = l1.next;
            } else {
                cur.next = new ListNode(l2.val);
                l2 = l2.next;
            }
            cur = cur.next;
        }
        return guard.next;
    }
}
